Problem 51 - University Physics with Modern Physics

Question

Starting with 2.50 mol of \(\mathrm{N}_{2}\) gas (assumed to be ideal) in a cylinder at 1.00 atm and \(20.0^{\circ} \mathrm{C},\) a chemist first heats the gas at constant volume, adding \(1.52 \times 10^{4} \mathrm{J}\) of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. (a) Calculate the final temperature of the gas. (b) Calculate the amount of work done by the gas. (c) Calculate the amount of heat added to the gas while it was expanding. (d) Calculate the change in internal energy of the gas for the whole process.

Step-by-Step Solution

Answer

(a) 879.26 K, (b) 60.780 J, (c) 3.297 × 10⁴ J, (d) 3.27 × 10⁴ J.

Step 1: Calculate Initial Volume and Pressure

The initial pressure of the gas is given as 1.00 atm, and we have 2.50 mol of \(_2\) at 20.0°C (or 293.15K). Using the ideal gas law, \PV = nRT\, we can find the initial volume: \[V_0 = \frac{nRT}{P} = \frac{2.50 \times 0.0821 \times 293.15}{1.00} = 60.0 L\]

Step 2: Calculate Final Temperature After Constant Volume Heating

Given the heat added at constant volume, \Q_1 = 1.52 \times 10^4 \, J\. For diatomic gases like \mathrm{N}_2\, use \C_V = \frac{5}{2} R\. The temperature change is given by \Q = nC_V\Delta T\, where =2.50\, and \R = 8.314\, J/(mol\cdot K)\.\[\Delta T_1 = \frac{Q_1}{nC_V} = \frac{1.52 \times 10^4}{2.50 \times \frac{5}{2} \times 8.314} = 146.48 K\]Therefore, the final temperature after constant volume heating is: \[T_1 = 293.15 + 146.48 = 439.63 \, K\]

Step 3: Calculate Final Temperature After Constant Pressure Expansion

Now the gas expands to twice its initial volume at constant pressure. By the combined gas law \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \\ \Rightarrow \ T_2 = 2T_1 = 2 \times 439.63 \), so the final temperature is \[T_2 = 879.26 \, K\].

Step 4: Calculate the Work Done by the Gas

For expansion at constant pressure, work done by the gas is \ W = P(V_2 - V_1)\. Since \ V_2 = 2V_1 \, \ V_1 = 60.0L \, and \ 1 \text{ atm} = 101.3 \text{ J/L} \\[W = P \Delta V = 1 \times 101.3 (2 \times 60.0 - 60.0) = 60.780 \, J\].

Step 5: Calculate the Heat Added During Expansion

Using \ Q_2 = \Delta U + W\ and knowing \ C_P - C_V = R\, we can calculate the heat added during constant pressure expansion \ Q_2 = nC_PT\. Hence, \( \Delta U = nC_V \Delta T_2 \) and we find:\[Q_2 = nC_PT_2 = \left(2.50 \times \frac{7}{2} \times 8.314 \times 439.63\right) = 3.297 \times 10^4 J\].

Step 6: Calculate the Internal Energy Change for the Entire Process

The change in internal energy \( \Delta U_{total}\) is given by the sum of the internal energy changes for each step:\[\Delta U_{total} = nC_V(T_2 - T_0) = 2.50 \times \frac{5}{2} \times 8.314 \left(879.26 - 293.15\right)\ \= 3.27 \times 10^4 \, J\].This step sums up the energy change from both heating phases.

Key Concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It enables us to understand how particles within a system, such as an ideal gas in a cylinder, react when they are subjected to changes in temperature, volume, and pressure. The exercise we're looking at involves the ideal gas law, which is central to thermodynamics. This law is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.

In the scenario provided, the chemist heats the gas, first at constant volume which increases its temperature, and then allows it to expand at constant pressure, causing further changes in temperature and volume. Understanding how an ideal gas behaves under these conditions is crucial in thermodynamics, as this behavior can be predicted and calculated using the ideal gas law and other related principles.

Heat Transfer

Heat transfer is the process of energy movement from one body or material to another, influenced by a temperature difference. In this exercise, heat is added to the nitrogen gas initially at a constant volume. This heat transfer increases the internal energy of the gas, which is absorbed and results in an increase in temperature. When heat is applied at constant volume, the relationship \( Q = nC_V\Delta T \) is used, where \( C_V \) is the molar heat capacity at constant volume.

Later in the exercise, heat is again transferred to the gas during its expansion at constant pressure. At constant pressure, heat transfer can be calculated using the equation \( Q_2 = nC_PT \), where \( C_P \) is the molar heat capacity at constant pressure. Heat transfer, both at constant volume and constant pressure, plays a critical role in determining the changes in energy within cyclic thermodynamic processes such as this one.

Internal Energy Change

Internal energy change \( (\Delta U) \) is a fundamental concept in thermodynamics and refers to the total change in energy within a system. For an ideal gas, changes in internal energy are directly related to the changes in the temperature of the gas. This is calculated using \( \Delta U = nC_V(T_2 - T_0) \), where \( T_0 \) and \( T_2 \) are the initial and final temperatures, respectively.

In the given exercise, the internal energy change for the whole process involves two phases: first, when the gas is heated at a constant volume and second, when it is heated again during expansion at constant pressure. The total internal energy change is obtained by summing the changes occurring during each phase. This concept helps determine the amount of energy absorbed or released during a thermodynamic process and is vital for calculating efficiency and understanding the behavior of thermodynamic systems.